Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
B(c(x1)) → B(a(c(x1)))
D(x1) → C(x1)
D(x1) → B(c(x1))
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
B(c(x1)) → B(a(c(x1)))
D(x1) → C(x1)
D(x1) → B(c(x1))
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(x1)
C(c(c(a(x1)))) → D(d(x1))
D(x1) → C(x1)
D(b(x1)) → C(x1)
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C(c(c(a(x1)))) → D(x1)
D(x1) → C(x1)
D(b(x1)) → C(x1)
The remaining pairs can at least be oriented weakly.

C(c(c(a(x1)))) → D(d(x1))
D(b(x1)) → C(c(x1))
Used ordering: Polynomial interpretation [25]:

POL(C(x1)) = 4·x1   
POL(D(x1)) = 12 + 4·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 3 + x1   
POL(c(x1)) = 6 + x1   
POL(d(x1)) = 9 + x1   

The following usable rules [17] were oriented:

c(x1) → a(a(x1))
b(c(x1)) → b(a(c(x1)))
c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
d(x1) → b(c(x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(x1)))) → D(d(x1))
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(c(a(x1)))) → D(d(x1)) at position [0] we obtained the following new rules:

C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))
C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))
C(c(c(a(b(x0))))) → D(c(c(x0)))
C(c(c(a(x0)))) → D(b(c(x0)))
D(b(x1)) → C(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.
We have obtained the following QTRS:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

The set Q is empty.
We have obtained the following QTRS:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
C(c(c(a(b(x))))) → D(c(c(x)))
C(c(c(a(x)))) → D(b(c(x)))
D(b(x)) → C(c(x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D1(x)
C1(b(x)) → A(b(x))
B(d(x)) → C1(x)
D1(x) → C1(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C1(b(D(x)))
C1(b(x)) → C1(a(b(x)))
D1(x) → B(x)
B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(d(x))
B(a(c(c(C(x))))) → C1(c(D(x)))
C1(x) → A(x)
B(D(x)) → C1(C(x))
C1(x) → A(a(x))
B(a(c(c(C(x))))) → C1(D(x))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
QDP
                              ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D1(x)
C1(b(x)) → A(b(x))
B(d(x)) → C1(x)
D1(x) → C1(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C1(b(D(x)))
C1(b(x)) → C1(a(b(x)))
D1(x) → B(x)
B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(d(x))
B(a(c(c(C(x))))) → C1(c(D(x)))
C1(x) → A(x)
B(D(x)) → C1(C(x))
C1(x) → A(a(x))
B(a(c(c(C(x))))) → C1(D(x))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(a(c(c(C(x))))) → C1(c(D(x)))
B(D(x)) → C1(C(x))
B(a(c(c(C(x))))) → C1(D(x))
The remaining pairs can at least be oriented weakly.

A(c(c(c(x)))) → D1(x)
C1(b(x)) → A(b(x))
B(d(x)) → C1(x)
D1(x) → C1(b(x))
A(c(c(C(x)))) → B(D(x))
A(c(c(C(x)))) → C1(b(D(x)))
C1(b(x)) → C1(a(b(x)))
D1(x) → B(x)
B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(d(x))
C1(x) → A(x)
C1(x) → A(a(x))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = 8·x1   
POL(C(x1)) = 1   
POL(C1(x1)) = 4·x1   
POL(D(x1)) = 2   
POL(D1(x1)) = 8·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2·x1   
POL(c(x1)) = 4·x1   
POL(d(x1)) = 8·x1   

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ QDPOrderProof
QDP
                                  ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(x)
A(c(c(c(x)))) → D1(d(x))
C1(b(x)) → A(b(x))
B(d(x)) → C1(x)
C1(x) → A(x)
C1(x) → A(a(x))
D1(x) → C1(b(x))
A(c(c(C(x)))) → C1(b(D(x)))
A(c(c(C(x)))) → B(D(x))
D1(x) → B(x)
C1(b(x)) → C1(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
QDP
                                      ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

A(c(c(c(x)))) → D1(x)
B(d(x)) → C1(c(x))
C1(b(x)) → A(b(x))
A(c(c(c(x)))) → D1(d(x))
B(d(x)) → C1(x)
C1(x) → A(x)
C1(x) → A(a(x))
D1(x) → C1(b(x))
A(c(c(C(x)))) → C1(b(D(x)))
D1(x) → B(x)
C1(b(x)) → C1(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


A(c(c(C(x)))) → C1(b(D(x)))
The remaining pairs can at least be oriented weakly.

A(c(c(c(x)))) → D1(x)
B(d(x)) → C1(c(x))
C1(b(x)) → A(b(x))
A(c(c(c(x)))) → D1(d(x))
B(d(x)) → C1(x)
C1(x) → A(x)
C1(x) → A(a(x))
D1(x) → C1(b(x))
D1(x) → B(x)
C1(b(x)) → C1(a(b(x)))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = 2   
POL(C1(x1)) = x1   
POL(D(x1)) = 4   
POL(D1(x1)) = 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2·x1   
POL(c(x1)) = 4·x1   
POL(d(x1)) = 8·x1   

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ QDPOrderProof
QDP
                                          ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(x)
A(c(c(c(x)))) → D1(d(x))
C1(b(x)) → A(b(x))
B(d(x)) → C1(x)
C1(x) → A(x)
D1(x) → C1(b(x))
C1(x) → A(a(x))
C1(b(x)) → C1(a(b(x)))
D1(x) → B(x)

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(d(x)) → C1(c(x))
A(c(c(c(x)))) → D1(x)
A(c(c(c(x)))) → D1(d(x))
B(d(x)) → C1(x)
D1(x) → C1(b(x))
D1(x) → B(x)
The remaining pairs can at least be oriented weakly.

C1(b(x)) → A(b(x))
C1(x) → A(x)
C1(x) → A(a(x))
C1(b(x)) → C1(a(b(x)))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = 2·x1   
POL(C1(x1)) = x1   
POL(D(x1)) = 1 + 2·x1   
POL(D1(x1)) = 2 + x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   
POL(d(x1)) = 3 + x1   

The following usable rules [17] were oriented:

b(D(x)) → c(C(x))
b(a(c(c(C(x))))) → c(c(D(x)))
c(b(x)) → c(a(b(x)))
a(c(c(C(x)))) → c(b(D(x)))
c(x) → a(a(x))
a(c(c(c(x)))) → d(d(x))
d(x) → c(b(x))
b(d(x)) → c(c(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ QDPOrderProof
                                        ↳ QDP
                                          ↳ QDPOrderProof
QDP
                                              ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
The TRS P consists of the following rules:

C1(b(x)) → A(b(x))
C1(x) → A(x)
C1(x) → A(a(x))
C1(b(x)) → C1(a(b(x)))

The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
b(a(c(c(C(x))))) → c(c(D(x)))
a(c(c(C(x)))) → c(b(D(x)))
b(D(x)) → c(C(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.
We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(c(c(a(x1)))) → d(d(x1))
d(b(x1)) → c(c(x1))
b(c(x1)) → b(a(c(x1)))
c(x1) → a(a(x1))
d(x1) → b(c(x1))

The set Q is empty.
We have obtained the following QTRS:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

Q restricted rewrite system:
The TRS R consists of the following rules:

a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))

Q is empty.